A quality analyst wants to construct a sample mean chart for controlling a packaging process. He knows from past experience that whenever this process is in control, package weight is normally distributed with a mean of 20 ounces and a standard deviation of two ounces. Each day last week, he randomly selected four packages and weighed each Day Weight (ounces) Monday 23 22 23 24 Tuesday 23 21 19 21 Wednesday 20 19 20 21 Thursday 18 19 20 19 Friday 18 20 22 20 If he uses upper and lower control limits of 22 and 18 ounces, what is his risk (alpha) of concluding this process is out of control when it is actually in control (Type I error)

Respuesta :

Using the normal distribution and the central limit theorem, it is found that his risk is of 0.0456.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

In this problem:

  • Mean of 20 ounces and standard deviation of 2 ounces, hence [tex]\mu = 20, \sigma = 2[/tex].
  • Sample of 4 packages, hence [tex]n = 4, s = \frac{2}{\sqrt{4}} = 1[/tex]

The risk is the probability of being more than 2 ounces from the mean, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Applying the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{2}{1}[/tex]

[tex]Z = 2[/tex]

The risk is P(|Z| > 2), which is 2 multiplied by the p-value of Z = -2.

  • Looking at the z-table, Z = -2 has a p-value of 0.0228.

0.0228 x 2 = 0.0456

The risk is of 0.0456.

For more on the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213